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ngn
12:50 PM
@GalenIvanov hi! so, there are multiple implementations of k. the original one is super fast and super expensive, used by traders
 
@ngn ok
 
ngn
the 32-bit version is gratis for non-commercial use: kx.com/download
it's known usually as part of kdb+, notable for its simplicity, compactness, and speed
 
I think I had downloaded it once
 
ngn
then, there's kona, mit-licensed, reasonably fast, maybe somewhat bloated. it follows an old version of the language
there's oK, written in js (so very slow), but has new features from k5 and k6, so it's quite suitable for golfing; mit-licensed
correction: kona is isc-licenced, not mit-
 
I see kona has a wiki, oK - a manual; that's good
 
ngn
12:57 PM
@GalenIvanov unfortunately there's very little information online about k itself. oK's manual is a good one
there's little information because kx sell k only as part of kdb, which has a slightly different language exposed to the user - q
 
@ngn so maybe I'll strat with oK
 
ngn
q is a thin syntactic layer on top of k, in which symbolic verbs have been replaced with english words
 
Hey everyone, any golfs you guys can see here {+/1,(!⌽⍳⍵)>×\⌽⍳⍵}? I'm trying to remove the parenthesis but I don't know if I can >.>
 
ngn
@GalenIvanov if you intend to use it mostly for golfing, i think that's a good choice
 
@GalenIvanov That's where I started too. The online interpreter is pretty good
 
1:00 PM
@J.Sallé can you take the reverse range part out of the function?
 
@ngn Than it's decided - I'll start with oK
 
@Quintec nope
Well I can assign it to a variable, which I've tried already, for no byte loss
 
ngn
@GalenIvanov oK also has a graphics api ("iKe") and some cool demos in which you can look at and play with the source code
 
But it wasn't helping me much
 
@ngn great! I'll look at it
 
ngn
1:09 PM
@J.Sallé both ! and ×\ are applied to the same thing, so you could use a train: (!>×\)⌽⍳⍵
 
@ngn yeah I knew there was a possible train there somewhere. Thanks!
Okay, now I'm trying to go tacit and I can't figure something out: 1⊥1,(!>×\)⌽⍳100 gives me the right answer, but (1⊥1,(!>×\)⌽⍳)100 doesn't. Dyalog says both group the same way, 1 ⊥ 1 , !> ×\ ⌽⍳ , so why are the results different?
 
You need all three of (!>×\)⌽⍳ to be applied monadically, however the rules of trains say that that wont happen
You could do (!>×\)∘⌽∘⍳
 
Ah, I think I get it.
Trains still boggle my mind.
 
1:46 PM
The day I understand trains will be a glorious day
 
 
4 hours later…
5:16 PM
@ngn hm, looks like a bounty might actually go there o_o
 

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